Problem 10 - College Physics

Question

A hockey puck with mass 0.160 \(\mathrm{kg}\) is at rest on the horizontal, frictionless surface of a rink. A player applies a force of 0.250 \(\mathrm{N}\) to the puck, parallel to the surface of the ice, and continues to apply this force for 2.00 \(\mathrm{s}\) . What are the position and speed of the puck at the end of that time?

Step-by-Step Solution

Answer

The puck's position is 3.125 meters, and its speed is 3.125 m/s after 2 seconds.

Step 1: Identify Given Values

We will start by identifying the given values in the problem. A mass of the puck: \( m = 0.160 \, \text{kg} \), a force applied: \( F = 0.250 \, \text{N} \), and the duration of force application: \( t = 2.00 \, \text{s} \). The initial velocity \( v_0 \) is 0 since the puck starts at rest.

Step 2: Calculate the Acceleration

Use Newton's second law to find acceleration. According to the law, \( F = ma \). Therefore, acceleration \( a = \frac{F}{m} = \frac{0.250 \, \text{N}}{0.160 \, \text{kg}} = 1.5625 \, \text{m/s}^2 \).

Step 3: Determine the Final Velocity

Use the equation of motion \( v = v_0 + at \) to calculate the final velocity. Since \( v_0 = 0 \), \( v = 0 + (1.5625 \, \text{m/s}^2)(2.00 \, \text{s}) = 3.125 \, \text{m/s} \).

Step 4: Calculate the Final Position

Use the equation \( s = v_0 t + \frac{1}{2} at^2 \) to find the final position. Substituting the known values, \( s = 0 \times 2 + \frac{1}{2} \times 1.5625 \times 2^2 = 3.125 \, \text{m} \).

Step 5: Summarize the Results

The final position of the puck is 3.125 meters, and the final speed is 3.125 meters per second after 2.00 seconds.

Key Concepts

Acceleration Calculation

When understanding the movement of objects, calculating acceleration is crucial. Acceleration represents how an object's velocity changes over time. To find it, we use Newton's Second Law which connects force, mass, and acceleration with the formula: \[ F = ma \] Where:- \( F \) is the force applied, - \( m \) is the mass of the object, - \( a \) is the acceleration.In the case of the hockey puck, we know the force applied is \( 0.250 \, \text{N} \), and the puck's mass is \( 0.160 \, \text{kg} \). By rearranging the formula to solve for acceleration, we get\[ a = \frac{F}{m} \] Substituting the given values gives us:\[ a = \frac{0.250}{0.160} \approx 1.5625 \, \text{m/s}^2 \]This tells us how fast the puck's velocity is changing, or accelerating, as it is pushed along the ice.

Equations of Motion

Equations of motion allow us to predict the future movement of an object given its initial conditions and the forces acting on it. These equations incorporate information about time, velocity, acceleration, and displacement. For the hockey puck, we can use these equations to determine both the speed and position of the puck after the force is applied.

• The formula for final velocity \( v \) is based on initial velocity \( v_0 \), acceleration \( a \), and time \( t \):\[ v = v_0 + at \]Since the puck starts from rest, \( v_0 = 0 \), so:\[ v = 1.5625 \, \text{m/s}^2 \times 2 \, \text{s} = 3.125 \, \text{m/s} \]
• The equation to find the position \( s \) is:\[ s = v_0 t + \frac{1}{2} a t^2 \]Plugging in the values:\\[ s = 0 \times 2 + \frac{1}{2} \times 1.5625 \times 2^2 = 3.125 \, \text{m} \]

Each of these calculations helps us understand the progression of the puck's movement over the course of the time it was subject to the force.

Force and Mass Relationship

The interplay between force, mass, and acceleration is foundational to Newton's Second Law. This relationship dictates how an object's velocity will change when a force is applied. Understanding this concept helps us predict how objects respond to different forces based on their mass.

• **Force**: Measured in newtons (N), it represents the push or pull exerted on an object.
• **Mass**: A scalar quantity that indicates how much matter is in an object, measured in kilograms (kg). It tells us how resistant an object is to acceleration when a force is applied.
• **Acceleration**: The rate of change of velocity that occurs when a force acts on mass. It's measured in meters per second squared (\( \text{m/s}^2 \)).

These elements are intertwined in the formula \( F = ma \). A larger force results in greater acceleration, but only if the mass remains the same. Likewise, if the same force is applied to objects of differing mass, the lighter object accelerates more. In our example, this relationship enabled us to precisely calculate the puck's acceleration and subsequent motion after the advent of force.